Question

1. Solve using quadratic formula

a) x^2 + 12x - 13 = 0

Answer 1

Here we go!

The quadratic formula: x = [-b ± √(b2 - 4ac)]/2a
a = 1 (understood coefficient)
b = 12
c = -13

`x= (-b\pm √(b^2-4ac) )/(2a) \\ x= (-12\pm √(12^2-4(1)(-13)) )/(2(1)) \\ x= (-12\pm √(144-(-52)) )/(2) \\ x=(-12\pm √(196) )/(2) \\ x=(-12\pm14 )/(2)`

Okay, so here, we split into 2 problems. If 14 were positive:

`x=(-12+14 )/(2) \\ x= (2)/(2) \\ x=1`
If 14 were negative:

`x=(-12-14 )/(2) \\ \\ x= \frac {-26}{2} \\ x=-14`

So,
`\left \{ {{x=1} \atop {x=-14}} \right.`